Question

Two short magnets of equal dipole moment $M$ are fastened from ends at an angle ${120}^{\circ }$ as given in the figure. One of the magnet is perpendicular to the direction of magnetic field $B$. Find the required work to rotate the magnet such that both magnets make equal angles with the magnetic field ?

• A. $MB(\frac{\sqrt{3}+1}{2})$
• B. $MB(\frac{\sqrt{3}-1}{2})$
• C. $MB(\frac{\sqrt{3}}{2}-1)$
• D. $MB(\frac{\sqrt{3}}{2}+1)$

Answer 1

Answer: (C), (D)

$MB(\frac{\sqrt{3}}{2}+1)$

As we know, that direction of magnetic moment in a magnet is from south to north pole.


Thus net magnetic moment of the given arrangement is

$M_{net}=\sqrt{M^2+M^2+(2M\cdot M\cos {120}^{\circ })}$

$M_{net}=M$

The required work done to rotate a bar magnet in a uniform magnetic field is given by

$W=M_{net}B(\cos {\theta }_1-\cos {\theta }_2)$

Here, $M_{net}=M;{\theta }_1={30}^{\circ }$

$W=MB(\cos {30}^{\circ }-\cos {\theta }_2)$

There are two scenario present in which both magnets make equal angles.

From above diagram,

${\theta }_2=0^{\circ }\text{or}{180}^{\circ }$

Therefore,

$W=MB(\cos {30}^{\circ }-\cos 0^{\circ })=MB(\frac{\sqrt{3}}{2}-1)$

And,
$W^{\prime }=MB(\cos {30}^{\circ }-\cos {180}^{\circ })=MB(\frac{\sqrt{3}}{2}+1)$

Therefore, options $\left(C\right)$ and $\left(D\right)$ are correct.